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If y is an integer is y^3 divisible by 9 [#permalink]
19 Feb 2012, 08:43

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

86%(01:44) correct
14%(00:49) wrong based on 57 sessions

If y is an integer is y^3 divisible by 9

(1) y is divisible by 4(2) y is divisible by 6

OG answer is B - reason if y is divisible by 3 then y is divisble by 9 y is divisible by 6 so y is divisible by 3 and y is dvisible by 9

How is this possible - statement 2 means that Y is divisible by 6. Now Y is an integer therefore to be evenly divisible by 6 - y must be a multiple of 6.Not all multiples of 6 are divisible by 9.eg if Y = 12 then y is divisible by 6 but not divisible by 9 if Y = 18 then y is divisible by 6 and also by 9 So My answer is E

Re: If y is an integer is y^3 divisible by 9 [#permalink]
19 Feb 2012, 09:27

1

This post receivedKUDOS

Expert's post

If y is an integer is y^3 divisible by 9?

Notice that we are asked whether y^3 is divisible by 9 not y itself.

(1) y is divisible by 4. Clearly insufficient: y can be 4 or 9*4.

(2) y is divisible by 6 --> since y is divisible by 6 then it's divisible by 3 too: y=3k. So, y^3 will definitely have more than two 3's and thus will be divisible by 9: y^3=(3k)^3=9*(3k^3). Sufficient.